This is a valid joint pdf.
I just want to know if X1 and X2 are dependent or independent rvs ? Why ?
Thank you for your help.
Is there a way of seeing this without computing the marginal density of X1 ?
PS: If you suggest that they are INdependent, how do you explain the fact that the joint pdf cannot be factored into seperate functions containing each rv alone ?
This is easy to see. The joint density $f_{X_1,X_2}$ is not a product $f_{X_1,X_2}(x_1,x_2)=g(x_1)h(x_2)$ hence $(X_1,X_2)$ is not independent.
To see that the identity $f_{X_1,X_2}(x_1,x_2)=g(x_1)h(x_2)$ is impossible, check for example that $$ f_{X_1,X_2}(1,1)f_{X_1,X_2}(2,2)\ne f_{X_1,X_2}(1,2)f_{X_1,X_2}(2,1). $$ Likewise, for every $x_1\ne x_2$, $$ f_{X_1,X_2}(x_1,x_1)f_{X_1,X_2}(x_2,x_2)\ne f_{X_1,X_2}(x_1,x_2)f_{X_1,X_2}(x_2,x_1), $$ since the ratio $$ \frac{f_{X_1,X_2}(x_1,x_1)f_{X_1,X_2}(x_2,x_2)}{f_{X_1,X_2}(x_1,x_2)f_{X_1,X_2}(x_2,x_1)} $$ is $$ \exp\left(-1+\frac{x_1}{2x_2}+\frac{x_2}{2x_1}\right)=\exp\left(-\frac12\left(\sqrt{\frac{x_1}{x_2}}-\sqrt{\frac{x_2}{x_1}}\right)^2\right)\lt1. $$ Note finally (but this is not necessary to solve the question) that $(X_1,X_2)$ coincides in distribution with $(8UV,4V)$, where $(U,V)$ are i.i.d. standard exponential random variables. Then a shortcut to disprove independence would be to note that $$ E(X_1X_2)-E(X_1)E(X_2)=16E(U)\mathrm{var}(V)\ne0. $$